Q. A child is sitting on a merry-go-round that is spinning. If the child moves closer to the center, what happens to the angular velocity of the merry-go-round?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the child moves closer to the center, the moment of inertia decreases, causing the angular velocity to increase to conserve angular momentum.
Q. A child is sitting on a merry-go-round that is spinning. If the child moves towards the center of the merry-go-round, what happens to the angular velocity of the system?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, thus the angular velocity increases to conserve angular momentum.
Q. A child is sitting on a merry-go-round that is spinning. If the child moves towards the center, what happens to the angular velocity of the merry-go-round?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
As the child moves towards the center, the moment of inertia decreases, and to conserve angular momentum, the angular velocity must increase.
Q. A child sitting at the edge of a merry-go-round throws a ball tangentially. What happens to the angular momentum of the system (merry-go-round + child + ball)?
A.Increases
B.Decreases
C.Remains constant
D.Becomes zero
Solution
Angular momentum of the system remains constant due to conservation of angular momentum.
Q. A particle is moving in a circular path with a radius of 2 m and a speed of 3 m/s. What is the angular momentum of the particle if its mass is 4 kg?
Q. A particle is moving in a straight line with a velocity v. If it suddenly starts moving in a circular path of radius r, what will be its angular momentum about the center of the circular path?
A.0
B.mv
C.mvr
D.mv^2/r
Solution
Angular momentum L = mvr, where v is the linear speed and r is the radius of the circular path.
Q. A particle is moving in a straight line with a velocity v. What is its angular momentum about a point O located at a distance r from the line of motion?
A.0
B.mv
C.mvr
D.mv^2
Solution
Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point O.
Q. A particle moves in a circular path with a radius r and a constant speed v. If the speed is doubled, what happens to the angular momentum of the particle?
A.It remains the same
B.It doubles
C.It quadruples
D.It halves
Solution
Angular momentum L = mvr; if v is doubled, L also doubles.
Q. A particle moves in a straight line with a velocity v. What is its angular momentum about a point P located at a distance d from the line of motion?
A.mv
B.mvd
C.mdv
D.0
Solution
Angular momentum L = mvr, where r is the perpendicular distance from the line of motion to point P.
Q. A particle of mass m is moving in a circular path of radius r with a constant speed v. What is the angular momentum of the particle about the center of the circle?
A.mv
B.mvr
C.mr^2
D.mv^2
Solution
Angular momentum L = mvr, where v is the linear speed and r is the radius.
Q. A planet orbits the sun in a circular path. If the radius of the orbit is doubled, what happens to the angular momentum of the planet if its speed remains constant?
A.Doubles
B.Halves
C.Remains the same
D.Quadruples
Solution
Angular momentum L = mvr, so if the radius is doubled and speed remains constant, angular momentum doubles.
